Solution -「BJWC 2018」「洛谷 P4486」Kakuro

$\mathcal{Description}$

  Link.

  有一个 $n\times m$ 的网格图，其中某些格子被主对角线划成两个三角形，称这样的格子为特殊格；初始时，除了一些障碍格，所有空格子和特殊格的两个三角形内都分别填上了数字。称一个网格合法，当且仅当：

• 对于每个特殊格左下方的三角形，若其不是障碍格，则其下方连续的空格子内数字之和为三角形内数字；
• 对于每个特殊格右上方的三角形，若其不是障碍格，则其右方连续的空格子内数字之和为三角形内数字。

  为了使网格合法，你可以以一定代价将某个格子内的数 $+1$ 或 $-1$，修改每个格子的代价是独立的，亦有一些格子不能修改。

  求最小代价。

  $n,m\le30$，每个空格子至少在一个左下非障碍的特殊格子下方，或在一个右上非障碍的特殊格子右方。

  我干嘛还要再概括一遍那么长的题意 qwq。

$\mathcal{Solution}$

   这种行和列和的限制有一种套路的网络流建图模型：代表行限制结点连向其限制格子的入点，格子入点连向格子出点用于体现一些代价之类的限制，最后格子出点连向限制该格子的代表列限制的结点。

  本题，$+1/-1$ 并不方便用边上的费用体现，对于一个初始数字为 $v$，修改代价为 $w$ 的空格子 $c$，考虑如下建边：

$$(c,c',[v,v],0)\\ (c',c,[0,f),w)\\ (c,c',[0,+\infty],w)$$

  其中 $(u,v,[l,r],w)$ 表示一条从 $u$ 到 $v$，流量限制为 $[l,r]$，费用为 $w$ 的边。发现我们通过构造第一条边“必选”使 $c$ 取到初始的数字 $v$，再通过回流的 $(c,c')$ 做到 $-1$（在可行流里，就会体现为一个 $c\rightarrow c'\rightarrow c$ 的回路），$+1$ 比较简单，不在赘述。

  剩下的就简单啦，所有可修改的行限制、列限制、空格子数字都可以用这种边的组合体现，建出图后跑有源汇上下界最小费用可行流即可。

  最小费用可行流就是把计算可行流时，求最大流的算法换成求最小费用最大流的算法，由于如此建图不存在正向负权边，所以正确性保证。

$\mathcal{Code}$

/* Clearink */

#include <queue>
#include <cstdio>
#include <cassert>

#define int long long
typedef std::pair<int, int> pii;

const int MAXN = 100, INF = 0x3f3f3f3f;
int n, m, type[MAXN + 5][MAXN + 5], cnt, id[MAXN + 5][MAXN + 5];
int deg[MAXN * MAXN * 2 + 10];
pii num[MAXN + 5][MAXN + 5], cost[MAXN + 5][MAXN + 5];

inline int imin ( const int a, const int b ) { return a < b ? a : b; }

struct MaxFlowCostGraph {
	static const int MAXND = MAXN * MAXN * 4 + 4, MAXEG = MAXN * MAXN * 100;
	int ecnt, head[MAXND + 5], S, T, bound, curh[MAXND + 5], d[MAXND + 5];
	bool inq[MAXND + 5];
	struct Edge { int to, flw, cst, nxt; } graph[MAXEG * 2 + 5];

	MaxFlowCostGraph (): ecnt ( 1 ) {}

	inline void link ( const int s, const int t, const int f, const int w ) {
		graph[++ecnt] = { t, f, w, head[s] };
		head[s] = ecnt;
	}

	inline Edge& operator [] ( const int k ) { return graph[k]; }

	inline void operator () ( int s, int t, const int f, const int w ) {
		#ifdef RYBY
			printf ( "%lld %lld ", s, t );
			if ( f == INF ) printf ( "INF " );
			else printf ( "%lld ", f );
			printf ( "%lld\n", w );
		#endif
		link ( s, t, f, w ), link ( t, s, 0, -w );
	}

	inline bool spfa () {
		static std::queue<int> que;
		for ( int i = 0; i <= bound; ++i ) d[i] = INF, inq[i] = false;
		d[S] = 0, inq[S] = true, que.push ( S );
		while ( !que.empty () ) {
			int u = que.front (); que.pop ();
			inq[u] = false;
			for ( int i = head[u], v; i; i = graph[i].nxt ) {
				if ( graph[i].flw && d[v = graph[i].to] > d[u] + graph[i].cst ) {
					d[v] = d[u] + graph[i].cst;
					if ( !inq[v] ) que.push ( v ), inq[v] = true;
				}
			}
		}
		return d[T] != INF;
	}

	inline pii dfs ( const int u, const int iflw ) {
		if ( u == T ) return { iflw, 0 };
		inq[u] = true; pii ret ( 0, 0 );
		for ( int& i = curh[u], v; i; i = graph[i].nxt ) {
			if ( graph[i].flw && !inq[v = graph[i].to]
			&& d[v] == d[u] + graph[i].cst ) {
				pii oflw ( dfs ( v, imin ( iflw - ret.first, graph[i].flw ) ) );
				graph[i].flw -= oflw.first, graph[i ^ 1].flw += oflw.first;
				ret.first += oflw.first;
				ret.second += graph[i].cst * oflw.first + oflw.second;
				if ( ret.first == iflw ) break;
			}
		}
		if ( !ret.first ) d[u] = INF;
		return inq[u] = false, ret;
	}

	inline pii calc ( const int s, const int t ) {
		S = s, T = t;
		pii ret ( 0, 0 );
		while ( spfa () ) {
			for ( int i = 0; i <= bound; ++i ) inq[i] = false, curh[i] = head[i];
			pii tmp ( dfs ( S, INF ) );
			ret.first += tmp.first, ret.second += tmp.second;
		}
		return ret;
	}
} graph;

inline void readKakuro ( pii arr[MAXN + 5][MAXN + 5] ) {
	for ( int i = 1; i <= n; ++i ) {
		for ( int j = 1; j <= m; ++j ) {
			arr[i][j].first = arr[i][j].second = -1;
			if ( type[i][j] == 1 || type[i][j] == 4 ) {
				scanf ( "%lld", &arr[i][j].first );
			} else if ( type[i][j] == 2 ) {
				scanf ( "%lld", &arr[i][j].second );
			} else if ( type[i][j] == 3 ) {
				scanf ( "%lld %lld", &arr[i][j].first, &arr[i][j].second );
			}
		}
	}
}

inline int ident ( const int i, const int j, const bool r, const bool t = false ) {
	assert ( 1 <= i && i <= n );
	assert ( 1 <= j && j <= m );
	assert ( type[i][j] && ( !t || ( t && ( type[i][j] == 2 || type[i][j] == 3 ) ) ) );
	return r * cnt + id[i][j] + t;
}

inline void specLink ( const int s, const int t, const int f, const int c ) {
	if ( ~c ) {
		// (s,t,[1,f],-c) and (s,t,[0,INF],c).
		graph ( s, t, INF, c );
		graph ( t, s, f - 1, c ), deg[s] -= f, deg[t] += f;
	} else {
		deg[s] -= f, deg[t] += f;
	}
}

signed main () {
	scanf ( "%lld %lld", &n, &m );
	for ( int i = 1; i <= n; ++i ) {
		for ( int j = 1; j <= m; ++j ) {
			scanf ( "%lld", &type[i][j] );
			if ( type[i][j] ) id[i][j] = ++cnt, cnt += type[i][j] < 4;
		}
	}
	int rS = cnt << 1 | 1, rT = rS + 1;
	int vS = rT + 1, vT = graph.bound = vS + 1;
	#ifdef RYBY
		printf ( "(%lld,%lld) & (%lld,%lld)\n", rS, rT, vS, vT );
	#endif
	readKakuro ( num ), readKakuro ( cost );
	for ( int i = 1; i <= n; ++i ) {
		for ( int j = 1; j <= m; ++j ) {
			if ( !type[i][j] ) continue;
			if ( type[i][j] == 4 ) {
				specLink ( ident ( i, j, 0 ), ident ( i, j, 1 ),
					num[i][j].first, cost[i][j].first );
				continue;
			}
			if ( ~num[i][j].first ) {
				int cur = ident ( i, j, 0, 0 );
				specLink ( rS, cur, num[i][j].first, cost[i][j].first );
				for ( int k = i + 1; type[k][j] == 4; ++k ) {
					graph ( cur, ident ( k, j, 0 ), INF, 0 );
				}
			}
			if ( ~num[i][j].second ) {
				int cur = ident ( i, j, 1, 1 );
				specLink ( cur, rT, num[i][j].second, cost[i][j].second );
				for ( int k = j + 1; type[i][k] == 4; ++k ) {
					graph ( ident ( i, k, 1 ), cur, INF, 0 );
				}
			}
		}
	}
	int req = 0;
	#ifdef RYBY
		puts ( "balancing degree..." );
	#endif
	for ( int i = 1; i <= rT; ++i ) {
		if ( deg[i] > 0 ) graph ( vS, i, deg[i], 0 );
		else if ( deg[i] ) req -= deg[i], graph ( i, vT, -deg[i], 0 );
	}
	graph ( rT, rS, INF, 0 );
	pii res ( graph.calc ( vS, vT ) );
	#ifdef RYBY
		printf ( "req = %lld;\nres = %lld %lld.\n", req, res.first, res.second );
	#endif
	if ( res.first != req ) puts ( "-1" );
	else printf ( "%lld\n", res.second );
	return 0;
}